Please check my solution of the following problem: Let $V_\lambda$ be an irreducible representation of Lie algebra $\mathfrak{sl}(2,\mathbb{C})$ with the highest weight $\lambda$. Decompose into a direct sum of irreducible representations $\mbox{Sym}^2(V_5)$.
Since $\lambda = 5$ is the highest weight, the character is given by $\chi_{V_5} = q^5 + q^3 + q^1 + q^{-1} + q^{-3} + q^{-5}$. $$(q^5 + q^3 + q^1 + q^{-1} + q^{-3} + q^{-5})^2 = q^{10} + 2q^8 + 3q^6 + 4q^4 + 5q^2 + 6 + 5q^{-2} + 4q^{-4} + 3q^{-6} + 2q^{-8} + q^{-10}$$
So the decomposition is $\mbox{Sym}^2(V_5) = V_{10} \oplus V_8 \oplus V_6 \oplus V_4 \oplus V_2 \oplus V_0$ and the dimensions of irreducible components are $\mbox{dim} V_{10} = 1$, $\mbox{dim} V_8 = 2$, $\mbox{dim} V_6 = 3$, $\mbox{dim} V_4 = 4$, $\mbox{dim} V_2 = 5$, $\mbox{dim} V_0 = 6$.
Is it right? It seems to me that I have done the task for $V_5 \otimes V_5$ but not for $\mbox{Sym}^2 V_5$.If so, please explain to me what to do with the symmetric power.